The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:
CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol
As observed O2 will not have enthalpy of formation as it is a pure substance.
To calculate for the enthalpy of reaction,
enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
= 294.5 kJ/mol
ANSWER: 294.5 kJ/mol
Answer:
3.87 x 10^24
Explanation:
Simply multiply the moles by avogadros number
6.42 moles of H2O x 6.022 x 10^23 molecules/1 mole of H2O = 3.87 x 10^24 molecules of H2O
Methane is a hydrocarbon which when burns in air (combustion) produces carbon dioxide and water. The equation for the reaction;
CH4 +2O2 = CO2 +2H2O
When one mole of methane combusts 2 moles of water are formed
Therefore; when 22 moles of methane combusts 44 moles of water are formed (22 ×2)
Answer:
B. the products have a smaller number of available energy microstates than the reactants.
The formula for kinetic energy is KE=1/2(mv²). Since both mass and velocity are multiplied by each other, particle with a larger mass needs to be moving slower than a particle with less mass if both have the same kinetic energy. You can think of it as 2KE/m=v² or 2KE/v²=m, If you increase the mass the velocity needs to decrease to keep the same KE value.
I hope this helps. Let me know in the comments if anything is unclear.
